A362677 - OEIS

%I #42 Dec 29 2024 14:09:03

%S 7,421,827,4733,32831,57571,228301,364751,892079,1932677,2256713,

%T 3684211,4213591,6751853,7218259,7887707,8497033,15720487,19925251,

%U 21055813,28756943,29547961,47369149,51881849,55033973,57954643,59677001,63062963,74415157,88535987

%N Primes whose reversal + 1 is a cube.

%C From _Jon E. Schoenfield_, Jul 03 2023: (Start)

%C Equivalently, primes whose reversal is one less than the cube of a positive integer whose last digit is not a 1.

%C Since no prime starts or ends with a 0, reversing the prime will not change the number of digits, and since no prime consists only of 9's, adding 1 to its reversal will not change the number of digits, either, so the cube will have the same number of digits as the prime. Since the prime cannot begin with a 0, its reversal cannot end in 0, so the cube cannot end in 1 (and a cube ends in 1 if and only if its cube root ends in 1). Since cubes are less dense than primes, a reasonably efficient but simple way to search for all terms having at most D digits is to test each positive integer r < 10^(D/3) such that r mod 10 != 1: if the reversal of r^3 - 1 is a prime, then that prime is a term of the sequence. (End)

%H Michael S. Branicky, <a href="/A362677/b362677.txt">Table of n, a(n) for n = 1..10000</a>

%e 421 is prime and reversal(421) + 1 = 124 + 1 = 125 = 5^3.

%e 364751 is in the sequence because it is prime and reversal(364751) + 1 = 157463 + 1 = 157464 = 54^3.

%t Select[Prime@Range@1000000,IntegerQ@CubeRoot@(FromDigits@Reverse@IntegerDigits@#+1) &]

%t r = Select[Range@300, Mod[#, 3] != 1 && Mod[#, 10] != 1 &];

%t s = Sort@Select[FromDigits /@ Reverse /@ IntegerDigits@(r^3 - 1),PrimeQ]

%o (Python)

%o from sympy import isprime

%o s=[int(str(k**3-1)[::-1]) for k in range(1,301)if k%10!=1 and k%3!=1]

%o t=[p for p in s if isprime(p)]

%o t.sort

%o print(t)

%Y Cf. A007488, A167217.

%K base,nonn,easy

%O 1,1

%A _Zhining Yang_, Jul 03 2023

%E a(18)-a(30) from _Jon E. Schoenfield_, Jul 03 2023